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求下列各三角函数值tan(-17π/6),tan(-31π/4)
人气:229 ℃ 时间:2020-05-13 11:17:54
解答
-17π/6=-18π/6 + π/6
=-3π +π/6
因此:
tan(-17π/6)=tan(-3π +π/6)=tan(-π+π/6)=tan(-5π/6)=-tan(5π/6)=-tan(π - π/6)=tan(π/6)
=√3/3
-31π/4=-32π/4+π/4=-8π+π/4
因此:
tan(-8π+π/4)=tanπ/4=1
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